If the speed of sound is 340 m/s, the two lowest frequencies of an 0.5 m organ pipe, closed at one end, are approximately:
A)170 and 340 Hz
B)170 and 510 Hz
C)340 and 680 Hz
D)340 and 1020 Hz
E)57 and 170 Hz

Question

Quizplus · Accepted Answer

The frequencies of a closed-end organ pipe can be found using the formula:

f = (2n-1)vs/4L

where:
f = frequency
n = harmonic number (1, 2, 3, ...)
v = speed of sound
s = pipe length (in this case, 0.5 m)
L = wavelength (in this case, since the pipe is closed at one end, the wavelength is four times the length of the pipe or 2m)

Plugging in the values, we get:

For the first harmonic (n=1):
f = (2n-1)vs/4L = (2(1)-1)340/(4(0.5)) = 170 Hz

For the second harmonic (n=2):
f = (2n-1)vs/4L = (2(2)-1)340/(4(0.5)) = 510 Hz

Therefore, the two lowest frequencies are approximately 170 and 510 Hz, and the best choice is B.

If the Speed of Sound Is 340 M/s, the Two